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(A) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth,while the lower half has length $l/2$ and is roug

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$2 	an \phi$

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(A) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth,while the lower half has length $l/2$ and is rough with coefficient of friction $\mu$.For the upper half (smooth):The acceleration is $a_1 = g \sin \phi$. The initial velocity $u = 0$. Using the equation $v^2 = u^2 + 2as$,the velocity $v$ at the midpoint is:$v^2 = 0 + 2(g \sin \phi)(l/2) = gl \sin \phi$.For the lower half (rough):The initial velocity is $v$ (from the midpoint),and the final velocity is $0$ (at the bottom). The acceleration is $a_2 = g(\sin \phi - \mu \cos \phi)$. Using $v_f^2 = v_i^2 + 2a_2s$:$0 = v^2 + 2g(\sin \phi - \mu \cos \phi)(l/2)$.Substituting $v^2 = gl \sin \phi$:$0 = gl \sin \phi + gl(\sin \phi - \mu \cos \phi)$.Dividing by $gl$:$0 = \sin \phi + \sin \phi - \mu \cos \phi$.$2 \sin \phi = \mu \cos \phi$.Therefore,$\mu = 2 	an \phi$.

The upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

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